Leetcode: Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1        /        2   5      /\        3   4   6

The flattened tree should look like:

   1         2             3                 4                     5                         6

IF you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

/** * Definition for a binary tree node. * struct TreeNode {*     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode (int x): Val (x), left (NULL) , right (NULL) {}}; */class Solution {public:    void flatten (treenode* root) {        if (root = NULL)    return;    treenode* Rightnode = root->right;//Save Right child    treenode* tmp = root->left;    while (TMP && tmp->right)    {    tmp = tmp->right;    } Find the right child direct precursor        if (TMP)//If the left child is not empty    {    tmp->right = rightnode;//Let the direct precursor point to the saved right child    Root->right = root->left;//right pointer to left child    Root->left = null;//left hand pointer    to empty flatten (root->right);//treat left child the same way    }    Flatten (Rightnode);//The same treatment for the right child    };

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Leetcode: Flatten Binary Tree to Linked List